Factor the following expression: $3$ $x^2+$ $23$ $x+$ $40$
Answer: This expression is in the form ${A}x^2 + {B}x + {C}$ . You can factor it by grouping. First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(40)} &=& 120 \\ {a} + {b} &=& & & {23} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $120$ and add them together. The factors that add up to ${23}$ will be your ${a}$ and ${b}$ When ${a}$ is ${8}$ and ${b}$ is ${15}$ $ \begin{eqnarray} {ab} &=& ({8})({15}) &=& 120 \\ {a} + {b} &=& {8} + {15} &=& 23 \end{eqnarray} $ Next, rewrite the expression as ${A}x^2 + {a}x + {b}x + {C}$ $ {3}x^2 +{8}x +{15}x +{40} $ Group the terms so that there is a common factor in each group: $ ({3}x^2 +{8}x) + ({15}x +{40}) $ Factor out the common factors: $ x(3x + 8) + 5(3x + 8) $ Notice how $(3x + 8)$ has become a common factor. Factor this out to find the answer. $(3x + 8)(x + 5)$